Solve Area and Volume Problems - 7th Grade Math

A toy block is formed by joining two rectangular prisms. Prism 1 is $8\text{ cm} \times 3\text{ cm} \times 4\text{ cm}$ . Prism 2 is $4\text{ cm} \times 3\text{ cm} \times 4\text{ cm}$ . They are joined along a full $3\text{ cm} \times 4\text{ cm}$ face (so there is no overlap). What is the total volume of the combined block?
Answer: $144\text{ cm}^3$ Tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). For this toy block made of two joined prisms, calculate volumes separately: Prism 1 volume is 8×3×4=96 cm³, Prism 2 volume is 4×3×4=48 cm³, and since they're joined without overlap, total volume is 96+48=144 cm³. The key phrase "no overlap" means we simply add the volumes without any subtraction. Common error would be somehow subtracting or miscalculating individual volumes. Steps: (1) identify composite structure (two rectangular prisms joined), (2) note "no overlap" means simple addition, (3) calculate each volume (8×3×4=96, 4×3×4=48), (4) combine by addition (96+48=144), (5) verify units (volume in cm³). When solids are joined without overlap, their volumes simply add together.
A poster is shaped like a $10\text{ in}\times 7\text{ in}$ rectangle with a $3\text{ in}\times 4\text{ in}$ rectangle cut out of one corner for a logo space. What is the area of the poster that remains?
Answer: $58\text{ in}^2$ This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this poster, large rectangle 10 in ×7 in=70 in² minus cutout 3 in ×4 in=12 in², remaining 58 in². Common errors include adding cutout (70+12=82 in²), or wrong area (10×7=70, but cutout 3×4=12; mistake like 10×4=40). Steps: (1) identify composite structure (rectangle with cutout), (2) decompose into standard shapes (large minus small rectangle), (3) calculate each component (apply A=lw), (4) combine (subtract), (5) verify units (area in²). Decomposition choice: large-minus-small (valid).
A science class builds a triangular prism. The triangular base has base $6\text{ cm}$ and height $4\text{ cm}$ , and the prism length is $10\text{ cm}$ . What is the volume of the triangular prism?
Answer: $120\text{ cm}^3$ This problem tests solving volume problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). For a triangular prism, first find the triangular base area: A = (1/2) × base × height = (1/2) × 6 × 4 = 12 cm², then multiply by prism length: V = 12 × 10 = 120 cm³. The correct volume is 120 cm³. Common errors include forgetting the (1/2) in the triangle area formula (using 6 × 4 = 24 instead of 12), which would give volume 240 cm³, or confusing dimensions. Steps: (1) identify the shape (triangular prism), (2) calculate triangular base area using A = (1/2)bh = (1/2) × 6 × 4 = 12 cm², (3) multiply base area by prism length: V = 12 × 10 = 120 cm³, (4) verify units (cm³). The key is remembering that triangular prism volume = (triangular base area) × length, and the triangular area needs the (1/2) factor.
A small shipping box is a rectangular prism with dimensions 3 in x 4 in x 5 in. What is the total surface area of the box (including all 6 faces)?
Answer: 94 $in^2$ The box has three pairs of matching faces: two 3x4=12 faces, two 3x5=15 faces, and two 4x5=20 faces. Adding one of each pair and doubling gives the total surface area: 2(12+15+20) = 2(47) = 94 in squared, matching choice C. Choice A (60 in squared) is actually the box's volume (3x4x5), not its surface area. Choice B (47 in squared) adds the three face areas correctly but forgets to double the sum, and choice D (120 in squared) comes from miscounting or double-counting one of the face pairs.
A storage container is a composite solid: a rectangular prism with dimensions $5\text{ cm} \times 4\text{ cm} \times 6\text{ cm}$ and a rectangular pyramid on top that has the same $5\text{ cm} \times 4\text{ cm}$ base and height $3\text{ cm}$ . What is the total volume of the container?
Answer: $140\text{ cm}^3$ This problem tests solving volume problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). The container has a rectangular prism base: V₁ = 5 × 4 × 6 = 120 cm³, and a pyramid on top with base area B = 5 × 4 = 20 cm² and height 3 cm: V₂ = (1/3) × 20 × 3 = 20 cm³, so total volume = 120 + 20 = 140 cm³. The correct calculation gives 140 cm³. Common errors include forgetting the (1/3) factor for pyramid volume (calculating 20 × 3 = 60 instead of 20), or arithmetic mistakes like 120 + 20 = 130. Steps: (1) identify composite structure (prism plus pyramid), (2) calculate prism volume (5 × 4 × 6 = 120), (3) calculate pyramid volume using V = (1/3)Bh where B = 5 × 4 = 20 and h = 3, giving (1/3) × 20 × 3 = 20, (4) add volumes (120 + 20 = 140), (5) verify units (cm³). The pyramid formula V = (1/3)Bh is crucial—missing the (1/3) is a common mistake.
A cylindrical water tank has a radius of 6 feet and a height of 10 feet. Water is pumped out at a rate such that the water level drops 2 feet per hour. How many cubic feet of water are removed in the first 3 hours?
Answer: 216π cubic feet In 3 hours, the water level drops 2 × 3 = 6 feet. The volume of water removed is a cylinder with radius 6 feet and height 6 feet: V = πr²h = π(6²)(6) = 216π cubic feet. Choice B uses height of 10 instead of 6. Choice C doubles the correct answer. Choice D uses the total tank volume incorrectly.
A garden is shaped like an L. You can decompose it into two rectangles: one rectangle is $9\text{ m}\times 4\text{ m}$ and the other is $5\text{ m}\times 3\text{ m}$ . What is the total area of the garden?
Answer: $51\text{ m}^2$ This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this L-shaped garden, decompose into two rectangles: 9 m ×4 m=36 m² and 5 m ×3 m=15 m², total 51 m². Common errors include multiplying dimensions wrong (9×4=36, but 5×3=15, sum 51; mistake like 9×5=45 total), or assuming overlap and subtracting. Steps: (1) identify composite structure (L-shape), (2) decompose into standard shapes (two rectangles), (3) calculate each component (apply A=lw), (4) combine (add areas), (5) verify units (area m²). Decomposition choice: two rectangles (valid, no overlap assumed).
A school display is shaped like a "house": a rectangle with a triangle on top. The rectangle is $8\text{ in}$ wide and $5\text{ in}$ tall. The triangle on top has the same base as the rectangle ( $8\text{ in}$ ) and height $3\text{ in}$ . What is the total area of the display?
Answer: $52\text{ in}^2$ This problem tests solving area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). The house-shaped display has: rectangle area = 8 × 5 = 40 in², triangle area = (1/2) × 8 × 3 = 12 in², so total area = 40 + 12 = 52 in². The correct total area is 52 in². Common errors include forgetting the (1/2) in the triangle formula (using 8 × 3 = 24, giving total 64), or arithmetic mistakes in addition. Steps: (1) identify composite structure (rectangle with triangle on top), (2) calculate rectangle area (8 × 5 = 40), (3) calculate triangle area using A = (1/2)bh = (1/2) × 8 × 3 = 12, (4) add areas (40 + 12 = 52), (5) verify units (in²). The "house" shape is a common composite figure—remember the triangle on top uses the same base width as the rectangle below.
A rectangular box has dimensions $3\text{ in} \times 4\text{ in} \times 5\text{ in}$ . What is the total surface area of the box (all 6 faces)?
Answer: $94\text{ in}^2$ This problem tests solving surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). The box has 6 faces in 3 pairs: two 3×4 faces (area = 12 each), two 3×5 faces (area = 15 each), two 4×5 faces (area = 20 each), so total = 2(12) + 2(15) + 2(20) = 24 + 30 + 40 = 94 in². The correct surface area is 94 in². Common errors include counting only 3 faces instead of 6 (giving 47), or making arithmetic mistakes. Steps: (1) identify all 6 faces of the rectangular box, (2) calculate areas of the 3 different face types: 3×4=12, 3×5=15, 4×5=20, (3) multiply each by 2 (opposite faces), (4) sum: 2(12)+2(15)+2(20)=24+30+40=94, (5) verify units (in²). Remember that a rectangular box has 6 faces, not 3—each dimension pair creates 2 opposite faces.
An L-shaped classroom floor needs new carpet. The floor can be seen as a $10\text{ m} \times 8\text{ m}$ rectangle with a $4\text{ m} \times 3\text{ m}$ rectangular storage area cut out of one corner. What is the area of the carpeted floor?
Answer: $68\text{ m}^2$ This problem tests solving area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). The L-shaped floor is a 10m × 8m rectangle with a 4m × 3m cutout, so we calculate: large rectangle area = 10 × 8 = 80 m², cutout area = 4 × 3 = 12 m², carpeted area = 80 - 12 = 68 m². The correct answer is 68 m². Common errors include calculating only the large rectangle (80 m²) without subtracting the cutout, or making arithmetic mistakes like 80 - 12 = 56. Steps: (1) identify composite structure (rectangle with rectangular cutout), (2) calculate large rectangle area (10 × 8 = 80), (3) calculate cutout area (4 × 3 = 12), (4) subtract cutout from large rectangle (80 - 12 = 68), (5) verify units (m²). Alternative method: decompose into two rectangles that form the L-shape, which should give the same answer as a check.
A storage container is made by attaching two rectangular prisms side-by-side (no overlap). Prism 1 is $7\text{ ft}\times 3\text{ ft}\times 2\text{ ft}$ and Prism 2 is $4\text{ ft}\times 3\text{ ft}\times 2\text{ ft}$ . What is the total volume of the container?
Answer: $66\text{ ft}^3$ This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this container, decompose into two prisms: 7 ft ×3 ft ×2 ft=42 ft³ and 4 ft ×3 ft ×2 ft=24 ft³, total 66 ft³ (no overlap). Common errors include adding dimensions instead of volumes (7+4=11 ×3×2=66, coincidental), or wrong volume (7×3×2=42, 4×3×2=24, but sum 60 if arithmetic error). Steps: (1) identify composite structure (two attached prisms), (2) decompose into standard shapes (two prisms), (3) calculate each component (apply V=lwh), (4) combine (add volumes), (5) verify units (volume ft³). Mistakes: double-counting overlapping areas (but none here), arithmetic errors.
A triangular prism has a triangular base with base $9\text{ m}$ and height $4\text{ m}$ , and the prism length is $5\text{ m}$ . What is the volume of the prism?
Answer: $90\text{ m}^3$ This problem tests solving volume problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). For the triangular prism: triangular base area = (1/2) × 9 × 4 = 18 m², volume = 18 × 5 = 90 m³. The correct volume is 90 m³. Common errors include forgetting the (1/2) in the triangle area formula (using 9 × 4 = 36, giving volume 180), or confusing the prism length with other dimensions. Steps: (1) identify the shape (triangular prism), (2) calculate triangular base area using A = (1/2)bh = (1/2) × 9 × 4 = 18 m², (3) multiply base area by prism length: V = 18 × 5 = 90 m³, (4) verify units (m³). The triangular prism volume formula is (triangular base area) × length—don't forget the (1/2) factor in the triangle area.
A tent stake is shaped like a triangular prism. The triangular base has base $6\text{ cm}$ and height $4\text{ cm}$ , and the prism length is $10\text{ cm}$ . What is the volume of the tent stake?
Answer: $120\text{ cm}^3$ This question tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). Surface area: sum all face areas (rectangular prism 3×4×5 has faces: two 3×4=12, two 3×5=15, two 4×5=20, total: 2(12+15+20)=94). For this triangular prism, base area (1/2)×6 cm×4 cm=12 cm², then volume 12 cm² ×10 cm=120 cm³. Common errors include missing (1/2) for triangle (using 24 cm² ×10=240 cm³), or treating as rectangular prism (6×4×10=240 cm³). Steps: (1) identify composite structure (triangular prism), (2) decompose into base triangle and length, (3) calculate each component (apply formulas: A=(1/2)bh, V=base area × length), (4) combine (multiply), (5) verify units (volume cm³). Mistakes: forgetting factors (1/2 in formula), arithmetic errors, units wrong.
A small park is shaped like an L. You can decompose it into two rectangles: one rectangle is $9\text{ m} \times 4\text{ m}$ and the other is $5\text{ m} \times 3\text{ m}$ . What is the total area of the park?
Answer: $51\text{ m}^2$ The L-shaped park splits into two rectangles: one measuring 9 meters by 4 meters, with an area of 36 square meters, and another measuring 5 meters by 3 meters, with an area of 15 square meters. Adding these two areas together gives 36 plus 15, which equals 51 square meters, matching choice C. Choice B and choice D come from arithmetic errors when adding or multiplying the rectangle dimensions. Choice A mistakenly reports only the area of the larger rectangle without adding the smaller one. Decomposing an irregular shape into non overlapping rectangles and adding their areas is the key strategy for solving composite figure problems like this one.
A science class builds a triangular prism model. The triangular base has base $6\text{ in}$ and height $4\text{ in}$ . The length of the prism is $10\text{ in}$ . What is the volume of the triangular prism?
Answer: $120\text{ in}^3$ This question tests solving volume problems for prisms by applying formulas to the base and length. For a triangular prism, decompose by finding the triangular base area A = (1/2)bh, then V = base area × length; example: triangle with b=6 in, h=4 in (area 12 in²), length 10 in, volume 120 in³. Formulas include triangle A = (1/2)bh and prism V = base area × length. For example, base (1/2)×6×4=12, times 10=120; or another prism with base 12 in² and length 10 in giving 120 in³. The correct calculation is base area (1/2)×6×4=12 in², volume 12×10=120 in³. Common errors include forgetting (1/2) for triangle (using 24 in², volume 240 in³), using wrong dimensions, or arithmetic errors (12×5=60). Steps: (1) identify the prism and base triangle, (2) calculate base area with (1/2)bh, (3) multiply by length, (4) verify units in in³. Mistakes: missing the half in triangle area, confusing area with volume units, or double-counting.
A cardboard cutout is an irregular polygon that can be decomposed into a rectangle and two right triangles. - Rectangle: $8\text{ cm} \times 4\text{ cm}$ - Two congruent right triangles: each has base $4\text{ cm}$ and height $3\text{ cm}$ What is the total area of the cutout?
Answer: $44\text{ cm}^2$ This problem tests solving area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). The cutout has: rectangle area = 8 × 4 = 32 cm², each triangle area = (1/2) × 4 × 3 = 6 cm², two triangles total = 2 × 6 = 12 cm², so total area = 32 + 12 = 44 cm². The correct total area is 44 cm². Common errors include forgetting the (1/2) in triangle area formula (using 4 × 3 = 12 per triangle, giving total 56), or counting only one triangle instead of two. Steps: (1) identify composite structure (rectangle plus two triangles), (2) calculate rectangle area (8 × 4 = 32), (3) calculate one triangle area using A = (1/2)bh = (1/2) × 4 × 3 = 6, (4) multiply by 2 for two congruent triangles (2 × 6 = 12), (5) add all areas (32 + 12 = 44), (6) verify units (cm²).
A storage container is made by stacking a rectangular prism and a rectangular pyramid on top. The prism has dimensions $5\text{ ft} \times 4\text{ ft} \times 6\text{ ft}$ . The pyramid on top has the same base $5\text{ ft} \times 4\text{ ft}$ and height $3\text{ ft}$ . What is the total volume of the container?
Answer: $140\text{ ft}^3$ Tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). Formulas: triangle A=(1/2)bh, rectangle A=lw, rectangular prism V=lwh, pyramid V=(1/3)Bh (B=base area), triangular prism V=((1/2)bh)×length (triangle base area times prism length). For this composite container, calculate prism volume: V=5×4×6=120 ft³, then pyramid volume: V=(1/3)×(5×4)×3=(1/3)×20×3=20 ft³, so total volume is 120+20=140 ft³. Common error would be forgetting the (1/3) factor for pyramid volume, giving 120+60=180 ft³, or arithmetic mistakes. Steps: (1) identify composite structure (rectangular prism with pyramid on top), (2) decompose into standard shapes (prism and pyramid), (3) calculate each component (prism: 5×4×6=120, pyramid: (1/3)×5×4×3=20), (4) combine (120+20=140), (5) verify units (volume in ft³). The pyramid formula V=(1/3)Bh is crucial—without the (1/3), you'd get 180 ft³ instead of the correct 140 ft³.
A small park is shaped like a rectangle with two identical right triangles attached on the left and right sides. The rectangle is $6\text{ m}$ long and $4\text{ m}$ wide. Each triangle has base $4\text{ m}$ (the same as the rectangle's width) and height $3\text{ m}$ . What is the total area of the park?
Answer: $36\text{ m}^2$ Tests solving area, volume, and surface area problems for composite figures by decomposing into simpler shapes (rectangles, triangles, prisms, pyramids), applying formulas, and combining results. Composite figures: decompose into standard shapes (L-shape as two rectangles: 10×5=50 and 6×3=18, sum: 68; or as large minus cutout: 10×8=80 minus 4×3=12, difference: 68, equivalent). For this park, calculate rectangle area: A=6×4=24 m², then each triangle area: A=(1/2)×4×3=6 m², and since there are two identical triangles, total triangle area is 2×6=12 m², giving total park area 24+12=36 m². Common error would be forgetting the (1/2) in triangle formula (using 4×3=12 for each triangle, giving total 24+24=48) or counting only one triangle. Steps: (1) identify composite structure (rectangle with two triangles on sides), (2) decompose into shapes (one 6×4 rectangle, two 4×3 right triangles), (3) calculate each (rectangle: 6×4=24, each triangle: (1/2)×4×3=6), (4) combine (24+2×6=24+12=36), (5) verify units (area in m²). The key is remembering both the (1/2) factor for triangles and that there are two triangles to add.
A right triangular prism has a triangular base with legs of 5 cm and 12 cm, and the prism has a height of 8 cm. If the prism is cut by a plane parallel to its triangular base at a height of 3 cm from the bottom, what is the volume of the smaller piece?
Answer: 90 cubic centimeters The triangular base has area ½ × 5 × 12 = 30 sq cm. The smaller piece has the same base area but height of 3 cm. Volume = base area × height = 30 × 3 = 90 cubic cm. Choice B uses height of 4 cm (8-4 error). Choice C uses height of 5 cm (leg length confusion). Choice D uses height of 6 cm (8-2 error).
Use the table showing the dimensions of three different rectangular rooms. Room C measures 14 feet by 8 feet. If each room needs flooring that costs 4 dollars per square foot, and Room C also needs crown molding that costs 8 dollars per linear foot around its perimeter, what is the total cost for all materials needed for Room C?
Answer: $800 Room C measures 14 feet by 8 feet, so its floor area is 14 x 8 = 112 square feet, and its perimeter is 2 x (14 + 8) = 44 feet. The flooring costs 112 x $4 = $448, and the crown molding costs 44 x $8 = $352. Adding both material costs together gives $448 + $352 = $800 for all the materials needed, matching Choice B. Choice A only includes the flooring cost and leaves out the crown molding entirely. Choice C only includes the molding cost and leaves out the flooring. Choice D comes from using an incorrect perimeter of 36 feet instead of 44 feet.

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